线性系统的根轨迹分析法 (23).pdf
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1、1 Chapter 6 Compensation of Automatic Control Systems 6.1 Basic Concepts and Forms of Compensation 6.2 Phase Lead Compensation 6.3 Phase Lag Compensation 6.4 Lag-lead Compensation 6.5 PID Controller 2 1 1 Basic principle Lag-lead compensation The lag-lead compensation uses its lead part to increase
2、the phase margin of the system,and use its lag part to improve the steady-state performance of the system.This compensation method takes into account the advantages of phase lag compensation and phase lead compensation.It can not only accelerate the response speed and reduce the overshoot,but also s
3、uppress the high-frequency noise and improve the steady-state accuracy.3 2 2 Lag-lead compensator and its transfer function 1C)s(oU2R)s(iU2C1RTTssTTG skT sT s1)1,(1)(),(1)(1)1221c12The passive lag-lead compensation network.is the lag compensation part.T sT s1122T sT s1111is the lead compensation par
4、t.Lag-lead compensation 4 3 3 Distribution of zeros and poles of the lag-lead compensator T12T12Re Im 0 T sT sG skTTT sT s(1)(1)(),1,(1)(1)21c1212T11T1Lead compensation part.Lag compensation part.Lag-lead compensation 5 4 4 Bode plot of the lag-lead compensator T sTsG skTsT s(1)(1)(),1(1)(1)21c12Bod
5、e Diagram0-5-10-15-20)s/dar(/Bd/LT0.011T0.11T11T10190450-45-90-20T0.011)s/dar(/ged/T0.11T11T101TTk1010121Lag-lead compensation Assuming 6 5 5 Lag-lead compensation based on Bode plot e.g.The open-loop transfer function of the known unit feedback system is .Try to design a lag-lead compensator to mee
6、t the following requirements.s ssG sK(1)(2)()According to the requirement of static speed error coefficient,the following formula can be listed as s ssKsG ssKKssv(1)(2)2lim()lim1000K20Solution(1)the phase margin is 50and the gain margin is greater than or equal to10dB.(2)Static speed error coefficie
7、nt is 10/s.Lag-lead compensation Example 7 Draw the Bode plot of the system before compensation when K=20.10.00164216.02.01.040.020.0)s/dar(/)()s/dar(/Bd/)(L0164216.02.01.040.020.010.0072-081-09009-04-02-0020406g g=-32 Lg=-13dB The system is unstable before compensation.Example 8 10.00164216.02.01.0
8、40.020.0)s/dar(/)()s/dar(/Bd/)(L0164216.02.01.040.020.010.0072-081-09009-04-02-0020406 Select the magnitude crossover frequency of the compensated system.=1.5 We choose the compensated magnitude crossover frequency as 1.5rad/s,according to the requirement,the phase lead required here is 50.Example 9
9、 Determine the break frequency of the lag part in the lag-lead compensator.The parameter is determined by the maximum phase lead.T100.15rad/s11.52The larger break frequency is chosen as one tenth of the system magnitude crossover frequency after compensation,the parameter T2 can be obtained as 1sin1
10、mssss0.0151166.710.1516.6711When the parameter is equal to 10,m=54.9,according to the requirement,the phase lead required here is 50,so choose=10.Thus,another break frequency of the phase lag part is .T0.015rad/s12The transfer function of the lag part of the lag-lead compensator is .Example 10 Deter
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